Homework 2
Chapter 19
Problem 31. A d = 40.0 cm diameter loop is rotated in a uniform electric field until the position of maximum electric flux
is found. The flux in this position is measured to be ΦE = 5.20 · 105 N·m2 /C. What is the magnitude of the electric field?
ΦE = EA
(1)
5
E=
2
5.20 · 10 N·m /C
ΦE
ΦE
=
= 4.14 · 106 N/C
=
A
π(d/2)2
π · (0.200 m)2
(2)
Problem 36. An m = 10.0 g piece of Styrofoam carries a net charge of q = −0.700 µC and floats above the center of a large
horizontal sheet of plastic that has a uniform charge density σ on it’s surface. Find σ.
Because the Styrofoam is floating in equilibrium, the sum of forces in the vertical direction must be zero. So
σ
Fg = mg = FE = qE = q
20
20 mg
2 · 8.54 · 10−12 C2 /N·m2 · 0.0100 kg · 9.80 m/s2
σ=
=
= −2.39 · 10−6 C/m2
q
−0.700 · 10−6 C
(3)
(4)
Problem 55. Four identical point charges (q = +10.0 µC) are located on the corners of a rectangle as shown in Figure
P19.55. The dimensions of the rectangle are L = 60.0 cm and W = 15.0 cm. Calculate the magnitude and direction of the
resultant electric force exerted on the charge at the lower left corner by the other three charges.
θ
F32
F42
q1
q2
q4
q3
ĵ
W
î
L
F12
This is just a jazzed up version of Problem 15 from Recitation 1. The unit vector r̂ diagonally across from the upper right
is given by
r̂ = cos θî + sin θĵ
(5)
◦
◦
θ = arctan W/L + 180 = 194
(6)
cos θ = −0.970
(7)
sin θ = −0.243
(8)
So the electric field in the lower left corner is given by
X qi
q
q
q
r̂
=
k
(−
î)
+
(cos
θ
î
+
sin
θ
ĵ)
+
(−
ĵ)
E = ke
i
e
ri2
L2
(L2 + W 2 )
W2
i
1
cos θ
1
sin θ
= −ke q
−
î
+
−
ĵ
L2
L2 + W 2
W2
L2 + W 2
So the magnitude of E is given by
s
E = ke q
L−2 −
cos θ
2
L + W2
2
+
W −2
sin θ
− 2
L + W2
2
= 4.08 · 106 N/C
(9)
(10)
(11)
(Remembering to convert L and W to meters.) And the direction φ (measured counter clockwise from î) of E is given by
!
θ
−W −2 + L2sin
+W 2
φ = arctan
+ 180◦ = 263◦
(12)
θ
−L−2 + L2cos
2
+W
Where the +180◦ is because the tangent has a period of 180◦ , and the angle we want is in the backside 180◦ .
F = qE so the direction of F is the same as the direction of E. The magnitude of F is given by
F = 10.0 · 10−6 C · 4.08 · 106 N/C = 40.8 N
(13)